How do you sketch the angle whose terminal side in standard position passes through (-3,1) and how do you find sin and cos?

1 Answer

Please see below.

Explanation:

To sketch the angle in standard position,

One side of the angle is the positive #x# axis (the right side of the horizontal axis). That is the initial side.

The terminal side has one end at the origin (the point #(0,0)#, also the intersection of the two axes) and goes through the point #(-3,1)#. So locate the point #(-3,1)#. Starting at the origin and count #3# to the left and up #1#. That will get you to the point #(-3,1)#. Put a dot there.
Now draw a line from the origin through the point #(-3,1)#. Your sketch should look a lot like this:

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If we knew how the angle was made (which direction and how many times around the circle), we would show that also.

Give the angle a name. I will use #theta# (that is the Greek letter "theta")

Memorize this

If the point #(a,b)# lies on the terminal side of an angle in standard position, then let #r = sqrt(a^2+b^2)#

The angle has sine #b/r# and it has cosine #a/r#

For this question

We have #(a,b) = (-3,1)#, so

#r = sqrt((-3)^2+(1)^2) = sqrt(9+1) = sqrt10#.

So the sine of #theta# is

#sin(theta) = 1/sqrt10#

Many trigonometry teachers will insist that you write your answer with a rational number in the denominator. If this is something you have to do, multiply the answer by #sqrt10/sqrt10# to look like this:

#1/sqrt10 * sqrt10/sqrt10 = (1*sqrt10)/(sqrt10 * sqrt10) = sqrt10/10#

So we should answer

#sin(theta) = sqrt10/10#

The cosine of #theta# is #a/r# so we have

#cos(theta) = (-3)/sqrt10 = (-3)/sqrt10 * sqrt10/sqrt10 = -(3sqrt10)/10#

Our answer is
#cos(theta) = -(3sqrt10)/10#